= Here we have 3a2b3 × 3a2b3 = 32ab9a'b', and (7abc")3= 73ab3c=343abc, as required. 2612 Зас = 3c 3c Here we have (a — 22)2 = (3ac — 2b)3 9a2c2 (3c)2 1418 Ba3 - 12abe + 46", and (-4) = (- 1)" = 902 3 5. Given, (a + b)2 = a2 + 2ab + b2 and (a + b)3 = a3 + 3a2b + 3ab2 + b3, to find (a + b), or the fifth power of a + b. Here, (a + b)3 × (a + b)2 = (a + b)3 gives (a + b)3 : = (a3 + 3a2b+3ab+b3) × (a2 + 2ab+b2) = a3 + 5a1b + 10a3b2 + 10a2b3 + 5ab1 + b3, as required. (4.) Because the raising of compound or polynomial expressions to powers by common multiplication is often long. and very operose, we will now proceed to show how the process may generally be greatly abridged. To the end in view, we shall commence by giving the INVESTIGATION OF THE BINOMIAL THEOREM. 1. Supposing + h to stand for the binomial, and n for any positive whole number, then from (C), in the remarks under Ex. 8 (p. 48), in Multiplication, and from (A) in the remarks under Ex. 6 (p. 70), in Division, we get the equations,(b+h)" = bn + [(b + h)n−1 + (b + h)n−2b + (b + h)n−3f2 + .... + (b + h)bn−2 +bn−1] × h=b" + nb"¬1h + terms depending on h3, h3, etc., (1); and (b+h)−" = b¬"+[b¬~1(b+ h)−"+b¬2(b + h)−(n−1) + h)−(n-3) b-3(b + h)-(n-2) + b−4(b + h)~(~~3) + . . . . + b-"(b + h)−1] × — h = b―n — nb−(n+1)h + terms depending on h2, h3, etc., (2). Because n-1, n − 2, n − 3, etc., are positive whole numbers, it follows from the expansion in (1), that we shall have (b + h)n−1 = bn−1 + (n − 1)bn−2h + etc., (b + h)n−2b = [bn−2 + (n − 2)b”−3h + etc.] × b = bn−1 + (n − 2)bn−2h + etc., (b and so on; conse+ (n − 2) + (n − 3) + h)n−3f2 = bn−1 + (n − 3)bn−3h + etc., quently, from (1), we shall get [(n − 1) + (n − 4) + . . . . +1, making n 1 terms] b-2h2 = n(n − 1)¿n-272, as appears from the Appendix to Multiplica 1 2 *, tion and Division; consequently, we shall have (b + h)" bn + nbn¬1h + = 1-22+ etc. If we proceed with (2) b-n in the same way as before, we shall get (b + h)—” — b—” — n(n + 1) f − (n + 2) p2 — etc.; which can also clearly be obtained from the expansion of (1), by changing the sign of n. Hence, we shall have (b + h)"-1 = br−1 + (n − 1)b"—2h + (n − 1) (n 1) (n − 2)¿n-372 + etc., (b + h)n−27 = bn−1 + (n 1 bn-2h+ 2 (n-2) (n − 2) (n − 3)¿n—sp2 + etc., (b + h)”−372 = bn−1 + (n 1 quently, from (1) we shall get [("= + (n − 3) (n (n-1) (n-2) 1 2 n(n − 1) (n − 2) dix before cited) 1.2 3 xb"-33, which is the fourth term of the expansion of (b + h)"; and proceeding in like fourth term of the expansion of (b + h)-". Proceeding in like manner from the fourth terms of the expansions to get the fifth terms, and so on, it is clear that we shall have terms, (1); and (b + h)" b―n — nb−(n+1)h + = n(n + 1) 1 2 n(n + 1) (n + 2) b −(n+3)3 + etc., to an unlimited 1 2 3 number of terms, (2'); which are the required expansions of (1) and (2). 2. We will now investigate the theorem, when the exponent of the power is a positive or negative vulgar fraction. Putting a = b + h or a − b = h, and using to represent n m any vulgar fraction, it is clear that we shall have am- bm as is evident by dividing the numerator and denominator by = If the members of the equation are multiplied by a - b n n h, and bm added to the resulting products, and (b + h) is put for a in the result, we shall get (b + h)m = bm + n m to n M-1 m to m × h, (3). n If in (3) we put 1 for m, and then change n into it will become (b + h) = bm + [(b + h)m 1 + (b + 1)m~2b + (b + hym -372 + etc., to an unlimited number of terms] × h, (3′); which is clearly the same result that we shall get from the division of the numerator of the fraction in (3), by its de If we put h for ab, and b + h for a, the equation is easily changed to (b + h) ̄ m = b ̃m + n 1 (b + h) mb ̄m + (b + h) ̄mb If we put 1 for m, and change n into n b (b + h) ̄ m = b ̄m + [(b + h) ̄mb−2 + (b + h) ́ ̄ m-1 m n m m-1 m (5) will become +1b−2 + (b + h, (5'), which is clearly the same as to convert the fraction in (5) into a series. n Because bm is the first term of the expansion of (b+ h) b n-1 n in (3), it is clear that is the first term of the expansion -1 m of (b+ h), and that each term in the numerator of the n-1 n-1 fraction in (3) gives bm for the first term of its expansion, so that all the terms give nbm; and in a similar way, the terms of the denominator give mbm; consequently, (3) is h+terms depending on h3, h3, etc., (3"), as is evi m= n dent from the division of the numerator of the fraction by its denominator; and similarly, (5) gives (b + h) n-(+1) + terms in h2, h3, etc., (5"), which can at once be obtained from (3′), by changing the sign of Again, because the first terms of the (b + h)”−1, (b + h)"−2b, (b + h)−3b3, etc., are b1, b-1, bm-1, etc., we get from (3') (b + h)m = bm + (lm2 + bm-1+ bm + etc.) × h + etc., so that the coefficient of his expressed in an indeterminate form, when its true value is m -1 ; consequently, the conversion of (3) into (3') converts and so on, indefinitely; and it is clear that similar observations are applicable to (5) and (5′). etc., are the second terms of the expansions of (b + h)m -1 (b + hym~3⁄4b, (b + h)-b2, etc., in (3′), it follows that we shall have [(-1)+(-2)+(-3) + etc.]-12 = (since, by the Appendix to Multiplication and Division, 1) n /n m m bm-2h2, for the third term of the expansion of n (b + h); and by changing the sign of 22, it will become m n the third term of the expansion of (b + h). n Proceeding in a similar way, we readily find (1) m 3 n bm-33 for the fourth term of the expansion of n (b+ h)m; and by changing the sign of it will become m n the fourth term of the expansion of (b + h); and it is clear that the law of the terms of the expansions is the same as when the exponent of the power of the binomial is a positive or negative whole number. 3. If any positive or negative irrational number is represented by p, the expansion of (b + h) will be of the same general form as the expansions of (b + h)”, (b + h)−". For let n' denote any very small rational number, which has the same sign as p, then taking the series n', 2n', 3n', |